What are the divisors of 90?

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

There is a total of 12 positive divisors.
The sum of these divisors is 234.
The arithmetic mean is 19.5.

6 even divisors

2, 6, 10, 18, 30, 90

6 odd divisors

1, 3, 5, 9, 15, 45

How to compute the divisors of 90?

A number N is said to be divisible by a number M (with M non-zero) if, when we divide N by M, the remainder of the division is zero.

$N mod M = 0$

Brute force algorithm

We could start by using a brute-force method which would involve dividing 90 by each of the numbers from 1 to 90 to determine which ones have a remainder equal to 0.

$Remainder = N - (M \times ⌊ \frac{N}{M} ⌋)$

(where $⌊ \frac{N}{M} ⌋$ is the integer part of the quotient)

90 / 1 = 90 (the remainder is 0, so 1 is a divisor of 90)
90 / 2 = 45 (the remainder is 0, so 2 is a divisor of 90)
90 / 3 = 30 (the remainder is 0, so 3 is a divisor of 90)
...
90 / 89 = 1.0112359550562 (the remainder is 1, so 89 is not a divisor of 90)
90 / 90 = 1 (the remainder is 0, so 90 is a divisor of 90)

Improved algorithm using square-root

However, there is another slightly better approach that reduces the number of iterations by testing only integers less than or equal to the square root of 90 (i.e. 9.4868329805051). Indeed, if a number N has a divisor D greater than its square root, then there is necessarily a smaller divisor d such that:

$D \times d = N$

(thus, if $\frac{N}{D} = d$ , then $\frac{N}{d} = D$ )

90 / 1 = 90 (the remainder is 0, so 1 and 90 are divisors of 90)
90 / 2 = 45 (the remainder is 0, so 2 and 45 are divisors of 90)
90 / 3 = 30 (the remainder is 0, so 3 and 30 are divisors of 90)
...
90 / 8 = 11.25 (the remainder is 2, so 8 is not a divisor of 90)
90 / 9 = 10 (the remainder is 0, so 9 and 10 are divisors of 90)